Calculus of Vector-Valued Functions

Transcription

1 Chapter 3 Calculus of Vector-Valued Functions Useful Tip: If you are reading the electronic version of this publication formatted as a Mathematica Notebook, then it is possible to view 3-D plots generated by Mathematica from different perspectives. First, place your screen cursor over the plot. Then drag the mouse while pressing down on the left mouse button to rotate the plot. ü 3.. Vector-Valued Functions Students should read Section 3. of Rogawski's Calculus [] for a detailed discussion of the material presented in this section. A vector-valued function is a vector where the components of the vector are themselves functions of a common parameter (or variable). For example, r is a vector-valued function if rt = xt, yt, zt. If we think of t as the time variable, the rt describes the motion of a particle through three-dimensional space over time. What we want to do is to understand what path is taken. We do this through graphing in three dimensions. Also, sometimes it is helpful to consider the projections of these curves onto the coordinate planes. For example, the projection of rt on the xy-plane is xt, yt,. Example 3.. Trace the paths of each of the following vector functions and describe its projections onto the xy-, xz-, and yzplanes: a) rt = t, t,t b) rt = cos 3 t, sin 3 t, sin t Solution: We use the ParametricPlot3D command to trace the path of each curve and to see its projection. a) First, we look at the plot of rt = t, t,t: In[73]:= ParametricPlot3Dt, t, t, t, 3, 3, PlotStyle Red, ImageSize Out[73]= -5 This curve looks very much like a parabola in 3-D space. To see the projections, we look first at:

2 Chapter 3.nb In[74]:= ParametricPlot3Dt, t,, t, 3, 3, PlotRange,, PlotStyle Orange, ImageSize Out[74]= This is clearly a parabola in the xy-plane. In[75]:= ParametricPlot3Dt,, t, t, 3, 3, Ticks Automatic,,,, Automatic, PlotStyle Orange, ImageSize 5, ImagePadding 5, 5, 5, Out[75]= -5 And this clearly a line in the xz-plane.

3 Chapter 3.nb 3 In[76]:= ParametricPlot3D, t, t, t, 3, 3, Ticks,,, Automatic, Automatic, PlotStyle Orange, ImageSize Out[76]= -5 This last plot is also clearly a parabola, but in the yz-plane. b) Next, we look at rt = cos 3 t, sin 3 t, sin t: In[77]:= ParametricPlot3DCost 3, Sin t 3, Sin t, t,,, PlotStyle Orange, ImageSize Out[77]= Note that since both sine and cosine are periodic with period p, it is not necessary to extend the domain beyond - p or + p. The projection in the xy-plane is:

4 4 Chapter 3.nb In[78]:= ParametricPlot3DCost 3, Sin t 3,, t,,, PlotPoints, ImageSize Out[78]= The projection in the xz-plane is: In[79]:= ParametricPlot3DCost 3,,Sin t, t,,, ImageSize Out[79]= Lastly, the projection in the yz-plane is:

5 Chapter 3.nb 5 In[8]:= ParametricPlot3D, Sin t 3, Sin t, t,,, ImageSize Out[8]= Note that the last two projections are almost exactly alike. This is to be expected because the sine and cosine functions have the same graph, but p radians apart. ü Exercises In Exercises through 3, graph rt and its three projections onto the coordinate planes.. rt = cos t, cos t, sin t. rt = t + 5, e 8 t cos t, e 8 t sin t 3. rt = t, t, 5t + t 4. Which of the following curves have the same projection onto the xz-plane? Graph the three projections to check your answer. a. r t = t, e t, t b. r t = e t, t, t c. r 3 t = t, cos t, t ü 3.. Calculus of Vector-Valued Functions Students should read Section 3. of Rogawski's Calculus [] for a detailed discussion of the material presented in this section. Since vector-valued functions are differentiated and integrated component by component, Mathematica will handle this easily since it treats vectors as lists and automatically performs the indicated operation on each element of the list. The derivative of a vector valued function rt = xt, yt, zt is defined to be while the integral of rt is Similarly, the limit is defined by r' t = x' t, y' t, z' t rt t = xt t, yt t, zt t. lim tøa rt = lim tøa xt, lim tøa yt, lim tøa zt. Example 3.. Differentiate and integrate each of the following vector functions:

6 6 Chapter 3.nb a) rt = t, t,t b) st = cos 3 t, sin 3 t, sin t Solution: (a) In[8]:= In[8]:= In[84]:= Out[84]= Clearr, s, t rt_ : t, t,t st_ : Cost 3, Sint 3, Sin t t rt, t, In[85]:= rt t Out[85]= t, t3 3,t (b) In[86]:= Out[86]= t st 3 Cost Sint, 3 Cost Sint, Cos t In[87]:= st t Out[87]= 3 Sint 4 3 Cost Sin3 t, 4 Cos3 t, Cos t Limits are handled the same way both in the calculus of vector-valued functions and in Mathematica: Example 3.3. Evaluate limit hø rt+h-rt h for rt = t, t,t. Solution: Since rt has been defined in the previous example, we merely evaluate In[88]:= Out[88]= rt h rt Limit,h h, t, As we would expect, this limit gives us the same answer for r' t as in the previous example. Example 3.4. Evaluate limit t,4t, tø3 Solution: t 3.

7 Chapter 3.nb 7 In[89]:= Limitt, 4 t,, t t3 Out[89]= 4, 8, 8 Derivatives of Dot and Cross Products Using the formulas of the derivative of the dot and cross products for vector-valued functions is simple in Mathematica. As a reminder, the formulas are: d rt st = rt s' t + r' t st and d rtμst = rtμs' t + r' tμst dt dt Example 3.5. Evaluate d dt rt st and d dt rtμst for rt = t, t,t and st = cos 3 t, sin 3 t, sin t. Solution: In[9]:= Out[9]= In[9]:= Out[9]= t rt.st Cost 3 4 t Cos t 3 t Cost Sint 3t Cost Sint t Sint 3 Sin t t rtst t Cos t 6 t Cost Sint Sint 3 t Sin t, Cost 3 t Cos t 6 t Cost Sint Sin t, t Cost 3 3t Cost Sint 3 t Cost Sint Sint 3 Tangent Lines Example 3.6. Find the vector parametrization of the tangent line to rt = - t,5t, t 3 at the point t = and plot it along with rt. Solution: Recall that the tangent line at t has vector parametrization Lt = rt + t r' t : In[9]:= rt_ t,5t,t 3 r't Lt_ r t r' Out[9]= Out[93]= Out[94]= t,5t,t 3 t,5,3t t,5 5t, 3t Here is a plot of the curve and the tangent line.

8 8 Chapter 3.nb In[95]:= ParametricPlot3Drt, Lt, t,, 4, ImageSize Small - 3 Out[95]= NOTE: Recall that the plot can be rotated to better view it from different perspectives. ü Exercises In Exercises and evaluate the limits. lim tøp sin t, cos t, tan 4 t. lim tø, et - t+ t In Exercises 3 and 4 compute the derivative and integral. 3. rt = tan t, 4t -, sin t 4. rt = e t, e t 5. Find a parametrization of the tangent line at the point indicated and plot both the vector-valued curve and the tangent line on the same set of axes. 6. Evaluate d dt rgt for rt = 4 sin t, cos t and gt = t.,4t ü 3.3. Arc Length Students should read Section 3.3 of Rogawski's Calculus [] for a detailed discussion of the material presented in this section. The arc length of a path rt = x t, y t, z t for a t b is given by L = a b r' t t = a b x' t + y' t + z' t t and like the one-dimensional version is difficult to evaluate by hand. Thus Mathematica is the perfect tool for calculating this. Example 3.7. Compute the arc length of rt = - t,5t, t 3 over the interval t. Solution:

9 Chapter 3.nb 9 In[96]:= rt_ : t, 5 t, t 3 L Normr't t Out[97]= EllipticE ArcSinh 3 3, EllipticE ArcSinh 3 3, EllipticE ArcSinh 6 6, EllipticE ArcSinh 6 6, EllipticF ArcSinh 3 3, 4 4 EllipticF ArcSinh 3 3, EllipticF ArcSinh 6 6, 4 4 EllipticF ArcSinh 6 6, Note that the above output indicates that Mathematica cannot find an antiderivative for the integrand, and thus we need to find another technique to evaluate this integral. Hence, we next try the numerical integrate command, NIntegrate, which does give us our result: In[98]:= L NIntegrateNormr't, t,, Out[98]= 5.85 Speed The vector r' t is also known as the velocity vector as it points in the (instantaneous) direction of motion described by rt. Its length or norm, r' t, gives the speed at time t.

10 Chapter 3.nb Example 3.8. Compute the speed of rt = - t,5t, t 3 when t =,.5, and. Solution: The following output gives a list of speeds of r' t at the three given times using the Norm command, which calculates the norm of a vector: In[99]:= rt_ : t, 5 t, t 3 Speed Normr', Normr'.5, Normr' Out[]= 65, 4.754, 67 In[]:= N Out[]= 8.66, 4.754, Observe that the speed is increasing as we move along the path of rt from t = to t =. This can be seen graphically by plotting the speed: In[]:= Normr't PlotNormr't, t,, Out[]= 5 4 Abst 36 Abst 4 5 Out[3]= NOTE: Observe how the Norm command inserts absolute values around each vector component in the formula for r' t, which seems redundant since each component is squared. This is done because in Mathematica vector components are allowed to be complex-valued, in which case absolute values are needed to refer to their magnitudes. ü Exercises In Exercises and, compute the length of curve over the given interval.. rt = sin t, 6t, cos t, -6 t 6. rt = t, 8t 3,3t, t In Exercises 3 and 4, find the speed of a particle moving along the curve rt at the given value of t. 3. rt = e t-,5t, 5t, t = 4. rt = sin t, cos 4 t, sin 6 t, t =p t 5. Compute st = r' u u for rt = t,t, t 3 and interpret Mathematica's result.

11 Chapter 3.nb 6. For rt = 4 t, - 3 t, 4 t, compute s t as in the previous exercise. Then use s t to find an arc length parametrization of rt, that is, find js = t, where j is the inverse of s t, and check to see that rj s has unit speed, that is, r' j s =. Lastly, plot rt and rj s and compare them. 7. Consider the helix rt = a sin t, a cos t, ct. a. Find a formula for the arc length of one revolution of rt. b. Suppose a helix has radius, height 5, and makes three revolutions. What is its arc length? t u 8. The Cornu spiral is defined by rt = xt, yt, where xt = sin u and yt = t u cos u. a. Plot the Cornu spiral over various intervals for t. b. Find a formula for its arc length along the interval -a t a, where a is a positive real number. c. What is its arc length in the limit as a Ø? ü 3.4. Curvature Students should read Section 3.4 of Rogawski's Calculus [] for a detailed discussion of the material presented in this section. Vector tools previously studied including arc length enables one to study the idea of curvature, which serves as a measure of how a curve bends, that is, the rate of change in direction of a curve. In arriving at a definition of curvature, consider a path in vector form and parametrized by rt = x t, y t, z t The parametrization is classified as regular if r' t for all values of t and for which r t is defined. Assume then that rt is regular and define the unit tangent vector in the direction of r' t, denoted T t, as follows: Tt = r' t r' t. This unit tangent vector T at any point enables us to determine the direction of the curve at that point, so one may define the curvature k (Greek letter kappa) at a point as k= dt ds = T' t r' t, which represents the magnitude of the rate of change in the unit tangent vector with respect to arc length. One denotes the vector dtds as the curvature vector. Its scalar length therefore measures curvature. For example, a straight line has k = (zero curvature) as one would expect. For a circle of radius r, we have k = r (reciprocal of r). This makes sense since a larger circle should have smaller curvature. In general, if we were to secure a circle, called the osculating circle, that best fits a curve at a specific point on the curve, then curvature of the curve at such a point should agree with the curvature of the osculating circle, that is, k= r Moreover, the radius r of this circle is called the radius of curvature. Note that the equations linking k and r illustrate their inverse relationship: k= r and r= k Example 3.9. Compute the curvature k for a circle of radius r defined by

12 Chapter 3.nb rt = r cos t, r sin t Solution: We first compute the unit tangent vector T using the formula Tt = r' t r' t : In[4]:= Clearr, T, t, In[5]:= Out[5]= rt_ Cost, Sint r't Tt_ r'tsimplifynormr't Cost, Sint Out[6]= Sint, Cost Out[7]= Sint Abs Cost Abs Sint, Cost Abs Cost Abs Sint Observe that in this output Mathematica is not able to reduce the expression inside the radical, which simplifies to r as a result of the fundamental trigonometric identity cos x + sin x =. This is due to the Norm command, which employs absolute values. To remedy this, we use the formula r' t = r' t r' t instead of the Norm command. In[8]:= Out[8]= Tt_ r'tsqrtsimplifyr't.r't Sint, Cost We then compute the curvature using the formula k= T' t r' t : In[9]:= SqrtSimplifyT't.T't Simplifyr't.r't Out[9]= Since the radius r is assumed to be positive, we conclude that k= r = = as expected. r r Example 3.. Compute the curvature k for the curve defined by f x = x at the point 3, 9. Solution: Observe that the graph of a function y = f x can be parametrized by x = t and y = f t and hence rt = t, f t. In this case the formula for curvature reduces to In[]:= Out[]= Clearr, t, f rt_ t, ft t, ft

13 Chapter 3.nb 3 In[]:= Out[]= Tt_ r't Sqrtr't.r't SqrtSimplifyT't.T't Simplifyr't.r't, f t f t f t Out[3]= f t f t 3 f '' x which is the same as k=. With f x = + f ' x x, we get 3 In[4]:= ft_ t Out[4]= t Out[5]= 4t 3 At x = t = 3, the curvature becomes In[6]:=. t 3 Out[6]= Here is a plot of the curvature along with the function. In[7]:= Plotft,, t,, Out[7]= Example 3.. Compute the curvature k and the radius of curvature r for the curve defined by

14 4 Chapter 3.nb rt = - t, t +, Solution: Again we begin by computing the unit tangent vector T: 3 t3 + at t =. In[8]:= Clearr, T, t, In[9]:= rt_ t, t^, 3 t^3 r't Tt_ r'tsqrtsimplifyr't.r't Out[9]= Out[]= t, t, t3 3, t, t Out[]=, t, t t t t We then compute the curvature using the same formula as in the previous example and evaluate it at t = : In[]:= SqrtSimplifyT't.T't Simplifyr't.r't. t Out[]= t 4 Out[3]= 8 9 Hence, the curvature k=89 at t = and the corresponding radius of curvature is r=k=98. Curvature Formula (Cross Product) There is an alternative formula for calculating the curvature of space curves that involves the cross product and eliminates the need to compute the unit tangent vector function: k= r'' t μ r' t r' t 3 = at μ vt vt 3 Example 3.. Compute the curvature kt and the radius of curvature for the helix defined by rt = cos t, sin t, t for any real number t. Solution: We first find the derivative of the unit tangent vector with respect to t. In[4]:= Clearr, T, t, rt_ Cost, Sint, t r't r''t Out[5]= Out[6]= Out[7]= Cost, Sint, t Sint, Cost, Cost, Sint,

15 Chapter 3.nb 5 In[8]:= Out[8]= t_ SqrtSimplifyCrossr''t, r't.crossr''t, r't SqrtSimplifyr't.r't 3 It follows that k= and r = for all values of t. Hence, our helix is a curve of constant curvature. ü Exercises In Exercises and, find r' t and Tt and evaluate T.. rt = 3 + t i t j + 9 t k. vt = sin t, cos t, 3. Use Mathematica to find the curvature function kx for y = cos x. Also plot kx for x. Where does the curvature assume its maximum value? 4. Determine the unit normal vectors to rt = t i + sin t j at t = p 4 and t = 3 p Determine the curvature of the vector-valued function rt = 3 + t i + 6 t j t k. 6. Find a formula for the curvature of the general helix rt = a cos t i + a sin t j + ctk. ü 3.5. Motion in Three Space Students should read Section 3.5 of Rogawski's Calculus [] for a detailed discussion of the material presented in this section. Recall that the velocity vector is the rate of the change of the position vector with respect to time while the acceleration vector represents the rate of change of the velocity vector with respect to time. Moreover, speed is defined to be the absolute value of the velocity vector. In short, we have the following: vt = r' t, st = vt and at = v' t = r'' t One can secure the velocity vector and the position function if the acceleration vector is known via integration. More specifically: vt = t au u + v where v represents the initial velocity vector and rt = t vu u + v t + r where r is the initial position. Example 3.. Find the velocity vector, the speed, and the acceleration vector for the vector-valued function rt = t 3 i + - t j + 4 t k at time t =. Solution:

16 6 Chapter 3.nb In[9]:= Clearr, v, s, a rt_ t^3, t, 4 t^ vt_ r't st_ Sqrtvt.vt at_ r''t v s a Out[3]= Out[3]= t 3, t, 4 t 3 t,, 8 t Out[3]= 64 t 9t 4 Out[33]= Out[34]= 6 t,,8 3,, 8 Out[35]= 74 Out[36]= 6,, 8 Thus, v = r' = 3 i - j + 8 k, s = 74, and a = 6 i + 8 k. Example 3.3. Find rt and vt if a t = t i +4 j subject to the initial conditions v = 3 i - j and r =. Solution: We first solve for vt by integrating at: In[37]:= Out[38]= Clearr, v, a at_ t, 4 vt_ Integrateau, u,, t v, v t, 4 Out[39]= t v, 4 t v Here, the constant vector of integration v = v, v = 3, - equals the initial velocity: In[4]:= Out[4]= Solvev 3,, v, v v 3, v Thus, vt = t i + 4 t j + 3 i - j. In[4]:= Out[4]= vt_ vt. v 3, v 3 t, 4t Next, we solve for rt by integrating vt:

17 Chapter 3.nb 7 In[4]:= rt_ Integratevu, u,, t r, r Out[4]= r 3t t3 6, r t t Again, the constant vector of integration r = r, r =, equals the initial position: In[43]:= Out[43]= Solver,, r, r r, r Hence, rt = t t i + t - t j. Components of Acceleration There are two components of acceleration: tangential and normal. More precisely, the acceleration vector a can be decomposed as a = a T T + a N N, where a T = d s is the tangential component and a N =k ds dt = is the normal component. dt = a v v Moreover, one has a T + a N = a so that a N = a -a T and a T = a -a N. Example 3.4. Determine the tangential and normal components of acceleration for the vector function r t = t 3, t, t. Solution: In[44]:= Clearr, v, s rt_ t^3, t^, t r't r''t a v v Out[45]= Out[46]= Out[47]= In[48]:= t 3,t,t 3 t,t, 6 t,, speed SimplifySqrtr't.r't Out[48]= 4t 9t 4 The result in the last output represents the speed at time t. In order to secure the tangential component of the acceleration, we differentiate the previous output: In[49]:= Out[49]= at Dspeed, t 8t 36 t 3 4t 9t 4 The normal component of the acceleration is

18 8 Chapter 3.nb In[5]:= an r''t.r''t at Out[5]= 8 t 36 t t 4 4t 9t 4 In[5]:= Simplifyan Out[5]= 9t 9t 4 4t 9t 4 NOTE: The components of acceleration can also be found through the formulas a T = a v v Mathematica as follows: and a N = a v, confirmed using v In[5]:= at r''t.r'tsqrtr't.r't an SqrtCrossr''t, r't.crossr''t, r't Sqrtr't.r't Out[5]= 4t 8 t 3 4t 9t 4 Out[53]= 4 36 t 36 t 4 4t 9t 4 ü Exercises In Exercises and, calculate the velocity and acceleration vectors and the speed at the time indicated:. rt = t i + - t j + 5 t k, t =.. rt = cos t i + sin t j + tan t k, t = p Sketch the path rt = - t i + - t j for -3 t 3 and compute the velocity and acceleration vectors at t =, t =, and t =. 4. Find vt given at and the initial velocity v. a. at = t i + 3 j, v = i + j b. at = e t i + j + t + k, v = i - 3 j + k 5. Find rt and vt given at together with the initial velocity and position at rest: a. at = e 3 t i + 4 t j + t - k, v = i + j + k, r = i + 3 j + 4 k b. at = i + j + sin t k, v = i + j, r = i. 6. Find the decomposition of at into its tangential and normal components at the indicated point: a. rt = 3-4 t i + t + j + t k at t = b. rt = t i + e -t j + te -t k at t = 7. Show that the tangential and normal components of acceleration of the helix given by rt = cos t i + sin t j + t k are equal to and, respectively.